16.5: Calculation: Smoothing part

16.5.1: Calculation: $ \Big( F_{s} (\Xi_{s}) \Phi^{s,s+1}{\widehat F}_{s+1}(\times_{t=s+1}^n \Xi_{t}) \Big) $ in (16.9) (in $\S$16.3)


Put \begin{align} \widetilde{f}_{x_n}(\omega_n) & = \frac{1}{\sqrt{2 \pi} {q}_n} \exp[ -\frac{(x_n -c_n \omega_n - d_n)^2}{2{q}_n^2} ] \nonumber \\ & \approx \exp[ -\frac{(c_n \omega_n -( x_n-d_n))^2}{2{q}_n^2} ] \equiv \exp[ - \frac {1}{2} \Big(\widetilde{u}_n \omega_n -\widetilde{v}_n \Big)^2 ] \tag{16.19} \end{align}

where it is assumed that $c_n$, $d_n$ and $q_n$ are known $(t \in T)$. And thus, put

\begin{align} {\widetilde{u}_n} =\frac{c_n}{q_n} ,\quad {\widetilde{v}_n} = \frac{x_n - d_n}{q_n} \tag{16.20} \end{align}

And further, Lemma 16.3 implies that the causal operator $\Phi^{t-1,t}:L^\infty (\Omega_t) \to L^\infty ( \Omega_{t-1})$ is defined by

\begin{align} & f_{t-1}(\omega_{t-1}) = [\Phi^{t-1,t} \widetilde{f}_{x_t}](\omega_{t-1} ) \nonumber \\ \approx & \int_{- \infty}^{\infty} \frac{1}{\sqrt{2 \pi} r_t} \exp[ -\frac{(\omega_t - a_t \omega_{t-1} -b_t )^2}{2r_t^2} ] \exp[ -\frac{({\widetilde{u}}_t \omega_t-{\widetilde{v}_t})^2}{2} ] d \omega_{t} \nonumber \\ \approx & \exp[ -\frac{1}{2} \Big( \frac{\widetilde{v}_t}{\sqrt{1 +r_t^2 {\widetilde{u}}_t^2 }} - \frac{{\widetilde{u}}_t(a_t \omega_{t-1} + b_t)}{\sqrt{1 +r_t^2 {\widetilde{u}}_t^2}} \Big)^2 ] \approx \exp[ -\frac{1}{2} \Big({{u}}_{t-1} \omega_{t-1}-{{v}_{t-1}} \Big)^2 ] \tag{16.21} \end{align} where \begin{align} & u_{t-1}=-\frac{a_t {\widetilde u}_t}{\sqrt{1 +r_t^2 {\widetilde{u}}_t^2 }}, \quad v_{t-1}=\frac{ b_t {\widetilde u}_t - {\widetilde v}_t }{\sqrt{1 +r_t^2 {\widetilde{u}}_t^2 }} \tag{16.22} \end{align}

And also, Lemma 16.3 implies that \begin{align} \widetilde{f}_{x_{t-1}}(\omega_{t-1}) & = \exp[ -\frac{(c_{t-1} \omega_{t-1} + d_{t-1}-x_{t-1})^2}{2{q}_{t-1}^2} ] \exp[ -\frac{({{u}}_{t-1} \omega_{t-1}-{{v}_{t-1}})^2}{2} ] \nonumber \\ & \approx \exp [ -\frac{1}{2} ( \frac{c_{t-1}^2 + u_{t-1}^2 q_{t-1}^2}{q_{t-1}^2 } ) \Big(\omega_{t-1} - \frac{ {c_{t-1}(d_{t-1}-t_{t-1})} + {u_{t-1}v_{t-1}}{q_{t-1}^2} }{{c_{t-1}^2} + {u_{t-1}^2}{q_{t-1}^2}} \Big)^2 ] \nonumber \\ & \approx \exp[ -\frac{1}{2} \Big({\widetilde{u}}_{t-1} \omega_{t-1}-{\widetilde{v}_{t-1}} \Big)^2 ] \tag{16.23} \end{align} where \begin{align} {{\widetilde{u}}_{t-1}} = \frac{\sqrt{c_{t-1}^2 + u_{t-1}^2 q_{t-1}^2}}{q_{t-1} }, \;\; {{\widetilde{v}}_{t-1}} = \frac{ {c_{t-1}(d_{t-1}-t_{t-1})} + {u_{t-1}v_{t-1}}{q_{t-1}^2} }{q_{t-1} \sqrt{{c_{t-1}^2} + {u_{t-1}^2}{q_{t-1}^2}}} \tag{16.24} \end{align} Summing up the above (16.19)-(16.24), we see: \begin{align} { \overset{\widetilde{u}_s, \widetilde{v}_s}{ \underset{\widetilde{w}_s}{\fbox{$\widetilde{f}_{x_s}$}} }} \xleftarrow[]{\mbox{$x_s$}} \cdots \xleftarrow[]{\Phi^{t-2,t-1}} { \overset{\widetilde{u}_{t-1}, \widetilde{v}_{t-1}}{ \underset{\widetilde{w}_{t-1}}{ {{\fbox{$\widetilde{f}_{x_{t-1}}$}}}}}} \xleftarrow[(16.24)]{{x_{t-1}}} { \overset{{u}_{t-1}, {v}_{t-1}}{ \underset{w_{t-1}}{ \fbox{${f}_{t-1}$}}} } \xleftarrow[(16.22)]{\Phi^{t-1,t}} { \overset{\widetilde{u}_t, \widetilde{v}_t}{ \underset{{\widetilde{w}_t}}{ \fbox{$\widetilde{f}_{x_t}$}}} } \xleftarrow[]{{x_{t}}} \cdots \xleftarrow[]{{x_{n-1}}} { \overset{{u}_{n-1}, {v}_{n-1}}{ \underset{{w}_{n-1}}{ \fbox{${f}_{n-1}$}}} } \xleftarrow[]{\Phi^{n-1,n}} { \overset{{\widetilde{u}_n \widetilde{v}_n}}{\underset{{\widetilde{w}_n}}{\fbox{$\widetilde{f}_{x_n}$=(16.19)}}} } \end{align} And thus, we get \begin{align} \widetilde{f}_{x_s} \approx \lim_{\Xi_t \to x_t \;( t \in \{s.s+1, \cdots, n\})} \frac{ \Big( F_{s} (\Xi_{s}) \Phi^{s,s+1}{\widehat F}_{s+1}(\times_{t=s+1}^n \Xi_{t}) \Big)} { \| F_{s} (\Xi_{s}) \Phi^{s,s+1}{\widehat F}_{s+1}(\times_{t=s+1}^n \Xi_{t}) \Big) \|_{L^\infty (\Omega_s )} } \tag{16.25} \end{align} in (16.9) (in $\S$16.3)

After all, we solve Problem 16.2 (Kalman Filter), that is,
Answer 16.4 [The answer to Problem16.2 (Kalman Filter)]
$(A):$ Assume that a measured value $(x_0, x_2, \cdots, x_n )$ $(\in \times_{t=0}^n X_t)$ is obtained by the measurement ${\mathsf M}_{L^\infty (\Omega_0)}$ $(\widehat{\mathsf O}_{t_0},$ $ \overline{S}_{[\ast]}({{z}}_0 ) )$. Let $s(\in T)$ be fixed. Then, we get the Bayes-Kalman operator $[B_{\widehat{\mathsf{O}}_{t_0} }^s(\times_{t \in T} \{x_t\})] ({{z}}_0)$, that is, \begin{align} \Big([B_{\widehat{\mathsf{O}}_{t_0} }^s(\times_{t \in T} \{x_t\})] {{z}}_0 \Big)(\omega_s) = \frac{\widetilde{f}_{x_s}(\omega_s ) \cdot {{z}}_s (\omega_s )}{ \int_{-\infty}^{\infty} \widetilde{f}_{x_s}(\omega_s ) \cdot {{z}}_s (\omega_s ) d \omega_s } = z_s^a(\omega_s) \quad ( \forall \omega_s \in \Omega_s ) \end{align} where ${{z}}_s$ in (16.18) and $\widetilde{f}_{x_s}$ in (16.25) can be iteratively calculated as mentioned in this section.




Remark 16.5 The following classification is usual
$(B_1):$ Smoothing: in the case that $0 \le s < n$
$(B_2):$ Filter: in the case that $s= n$
$(B_3):$ Prediction: in the case that $s= n$ and, for any $m$ such that $n_0 \le m < n$, the existence observable $(X_m, {\mathcal F}_m, F_m )= (\{1\}, \{\emptyset ,\{1 \} \}, F_m )$ is defined by $F_m(\emptyset )\equiv 0$, $F_m(\{ 1 \} )\equiv 1$,